Perfect … Thanks Daniel! Paul Monday Parallel Scientific, LLC. paul.monday@parsci.com On Dec 8, 2011, at 2:13 PM, Daniel Fischer wrote:
On Thursday 08 December 2011, 21:55:46, Paul Monday wrote:
data UMatrix a = UMatrix (V.Vector (U.Vector a)) deriving (Show, Eq)
madd3 :: (U.Unbox a, Num a) => UMatrix a -> UMatrix a -> UMatrix a madd3 (UMatrix m) (UMatrix n) = UMatrix $ V.zipWith (U.zipWith (+)) m n
And it works … (really, when I started writing this note, I was completely lost … welcome to the power of sitting and walking through Haskell)
madd3 (UMatrix (Data.Vector.fromList [Data.Vector.Unboxed.fromList [1,2,3], Data.Vector.Unboxed.fromList [4,5,6], Data.Vector.Unboxed.fromList [7,8,9]])) (UMatrix (Data.Vector.fromList [Data.Vector.Unboxed.fromList [10,11,12], Data.Vector.Unboxed.fromList [13,14,5], Data.Vector.Unboxed.fromList [16,17,18]]))
I have ONE thing that I can't really explain …
Why does the declaration of madd3 REQUIRE the use of U.Unbox a as a data type? madd3 :: (U.Unbox a, Num a) => UMatrix a -> UMatrix a -> UMatrix a
I would have thought that my declaration of UMatrix would have done this?
No, the data declaration doesn't provide any such context, hence you have to specify it at the use site (I don't think for types which aren't instances of Unbox there are any values other than _|_, though).
You can make the Unbox constraint available from the data declaration if you use GADTs,
{-# LANGUAGE GADTs #-}
data UMatrix e where UMatrix :: U.Unbox a => V.Vector (U.Vector a) -> UMatrix a
Then pattern-matching on a UMatrix makes the Unbox dictionary of a available.